Problem: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}5x-3y &= 1 \\ 7x-y &= 1\end{align*}$
Explanation: Begin by moving the $y$ -term in the second equation to the right side of the equation. $7x = y+1$ Divide both sides by $7$ to isolate $x$ $x = {\dfrac{1}{7}y + \dfrac{1}{7}}$ Substitute this expression for $x$ in the first equation. $5({\dfrac{1}{7}y + \dfrac{1}{7}}) - 3y = 1$ $\dfrac{5}{7}y + \dfrac{5}{7} - 3y = 1$ Simplify by combining terms, then solve for $y$ $-\dfrac{16}{7}y + \dfrac{5}{7} = 1$ $-\dfrac{16}{7}y = \dfrac{2}{7}$ $y = -\dfrac{1}{8}$ Substitute $-\dfrac{1}{8}$ for $y$ in the top equation. $5x-3( -\dfrac{1}{8}) = 1$ $5x+\dfrac{3}{8} = 1$ $5x = \dfrac{5}{8}$ $x = \dfrac{1}{8}$ The solution is $\enspace x = \dfrac{1}{8}, \enspace y = -\dfrac{1}{8}$.